hẽhe kĩckDễ z sao đăg hả bn

\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}\)

\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge3+2=5\)

\(\Leftrightarrow5x\left(x+2\right)-6\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(5x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{6}{5}\end{matrix}\right.\)

a) \(3-x^2+5x=-\left(x^2-5x-3\right)\)

\(=-\left(x^2-2x.\frac{5}{2}+\frac{10}{4}-\frac{22}{4}\right)\)

\(=-\left(x-\frac{5}{2}\right)^2+\frac{22}{4}\)

\(=-\left(x-\frac{5}{2}\right)^2+\frac{11}{2}\)

Mà: \(\left(x-\frac{5}{2}\right)^2\ge0\)\(\Leftrightarrow-\left(x-\frac{5}{2}\right)^2\le0\)

\(\Leftrightarrow-\left(x-\frac{5}{2}\right)^2+\frac{11}{2}\le\frac{11}{2}\)

\(\Leftrightarrow3-x^2+5x\le\frac{11}{2}\)

Dấu = xảy ra khi: \(\left(x-\frac{5}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{5}{2}=0\)

\(\Leftrightarrow x=\frac{5}{2}\)(T/m)

Vậy GTLN của 3 - x² + 5x là \(\frac{11}{2}\)khi x = \(\frac{5}{2}\).

b) \(12-6x^2-6x=-6\left(x^2+x-2\right)\)

\(=-6\left(x^2+2x.\frac{1}{2}+\frac{1}{4}-\frac{9}{4}\right)\)

\(=-6\left(x+\frac{1}{2}\right)^2+\frac{27}{2}\)

Mà: \(\left(x+\frac{1}{2}\right)^2\ge0\)\(\Leftrightarrow-6\left(x+\frac{1}{2}\right)^2\le0\)

\(\Leftrightarrow-6\left(x+\frac{1}{2}\right)^2+\frac{27}{2}\le\frac{27}{2}\)\(\Leftrightarrow12-6x^2-6x\le\frac{27}{2}\)

Dấu = xảy ra khi: \(\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\frac{1}{2}=0\)\(\Leftrightarrow x=-\frac{1}{2}\)(T/m)

Vậy GTLN của 12 - 6x² - 6x là \(\frac{27}{2}\)khi x = \(-\frac{1}{2}\).

f) (2x - 8)(4x + 16) = 0

<=> 2x - 8 = 0 hoặc 4x + 16 = 0

<=> 2x = 0 + 8 hoặc 4x = 0 - 16

<=> 2x = 8 hoặc 4x = -16

<=> x = 4 hoặc x = -4

g) 5x(6x - 12) = 0

<=> 5x = 0 hoặc 6x - 12 = 0

<=> x = 0 hoặc 6x = 0 + 12

<=> x = 0 hoặc 6x = 12

<=> x = 0 hoặc x = 2

h) 7(9 - x)(12 - 6x) = 0

<=> 9 - x = 0 hoặc 12 - 6x = 0

<=> -x = 0 - 9 hoặc -6x = 0 - 12

<=> -x = -9 hoặc -6x = -12

<=> x = 9 hoặc x = 2

d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)

\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

\(=x^2-2x+1\)

\(=\left(x-1\right)^2\)

+) ta có : \(E=3x^2-6x+15=3\left(x^2-2x+1\right)+12\)

\(=3\left(x-1\right)^2+12\ge12\) \(\Rightarrow E_{min}=12\) khi \(x=1\)

+) ta có : \(F=5x^2+6x-12=5\left(x^2+\dfrac{6}{5}x+\dfrac{9}{25}\right)-\dfrac{69}{5}\)

\(=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{69}{5}\ge\dfrac{-69}{5}\) \(\Rightarrow F_{min}=-\dfrac{69}{5}\) khi \(x=\dfrac{-3}{5}\)

+) ta có : \(G=4x^2-4x+25=4\left(x^2-x+\dfrac{1}{4}\right)+24\)

\(=4\left(x-\dfrac{1}{2}\right)^2+24\ge24\) \(\Rightarrow G_{min}=24\) khi \(x=\dfrac{1}{2}\)

+) ta có : \(H=9x^2+6x^2+4=15x^2+4\ge4\)

\(\Rightarrow H_{min}=4\) khi \(x=0\)

Tìm GTNN

E=3x^2-6x+15

F= 5x^2+6x-12

G=4x^2-4x+25

H=9x^2+6x^2+4

`#3107`

`a)`

`(6x - 2)^2 + 4(3x - 1)(2 + y) + (y + 2)^2 - (6x + y)^2`

`= [(6x - 2)^2 - (6x + y)^2] + 4(3x - 1)(2 + y) + (2 + y)^2`

`= (6x - 2 - 6x - y)(6x -2 + 6x + y) + (2 + y)*[ 4(3x - 1) + 2 + y]`

`= (2 - y)(12x + y - 2) + (2 + y)*(12x - 4 + 2 + y)`

`= (2 - y)(12x + y - 2) + (2 + y)*(12x + y - 2)`

`= (12x + y - 2)(2 - y + 2 + y)`

`= (12x + y - 2)*4`

`= 48x + 4y - 8`

`b)`

\(5(2x-1)^2+2(x-1)(x+3)-2(5-2x)^2-2x(7x+12)\)

`= 5(4x^2 - 4x + 1) + 2(x^2 + 2x - 3) - 2(25 - 20x + 4x^2) - 14x^2 - 24x`

`= 20x^2 - 20x + 5 + 2x^2 + 4x - 6 - 50 + 40x - 8x^2 - 14x^2 - 24x`

`= - 51`

`c)`

\(2(5x-1)(x^2-5x+1)+(x^2-5x+1)^2+(5x-1)^2-(x^2-1)(x^2+1)\)

`= [ 2(5x - 1) + x^2 - 5x + 1] * (x^2 - 5x + 1) + (5x - 1)^2 - [ (x^2)^2 - 1]`

`= (10x - 2 + x^2 - 5x + 1) * (x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`

`= (x^2 + 5x - 1)(x^2 - 5x + 1) + (5x - 1)^2 - x^4 + 1`

`= x^4 - (5x - 1)^2 + (5x - 1)^2 - x^4 + 1`

`= 1`

`d)`

\((x^2+4)^2-(x^2+4)(x^2-4)(x^2+16)-8(x-4)(x+4)\)

`= (x^2 + 4)*[x^2 + 4 - (x^2 - 4)(x^2 + 16)] - 8(x^2 - 16)`

`= (x^2 + 4)(x^4 + 12x^2 - 64) - 8x^2 + 128`

`= x^6 + 16x^4 - 16x^2 - 256 - 8x^2 + 128`

`= x^6 + 16x^4 - 24x^2 - 128`